The Vertical Circle height h=2r acceleration due to gravity g Height x less than or equal to h   , x Put any  k value FIND velocity at B so that the particle does not fall off  at A:  u=√(5gr) vx=√(2gx) velocity at B so that the particle  at A is zero:  u1=√(2gr) velocity at A when velocity at B is u---  v=√(gr) velocity at A when velocity at B is u1----v1=0 radius of the circle r: velocity at P corresponding to u, vp=√(u2-2gx) velocity at P corresponding to u1,  vp1=√(u12-2gx) Angle between OP and OB  :θ in degree OX XP Path length BP Displacement BP Path length/displacement sinθ cosθ tanθ Time period of revolution=T threshhold Time period of revolution=T1 i.e particle shall reach the top and then free fall to ground. Frequency f  : x2 = vx2/u2 ; y2 = vp2/u2 ; x2+y2 =1 x y Let velocity at B be u2=√(5+x)gr The trajectory shall be a circle of radius r2=(5+x)r/5=(1+0.2*k)r velocity at A when velocity at B is u2---is  v2=√(gr2) When ground velocity is changed , the radius of the circle changes so that the apex velocity shall remain √(gr') and ground velocity is √(5gr'). When we revolve an object attached to a string, with increasing velocity, speed also increases at every point and the corresponding force is balanced by string tension till the string snaps.